PRoblem Description Given two rectangles and the coordinates of two points on the diagonals of each rectangle,you have to calculate the area of the intersected part of two rectangles. its sides are parallel to OX and OY .
Input Input The first line of input is 8 positive numbers which indicate the coordinates of four points that must be on each diagonal.The 8 numbers are x1,y1,x2,y2,x3,y3,x4,y4.That means the two points on the first rectangle are(x1,y1),(x2,y2);the other two points on the second rectangle are (x3,y3),(x4,y4).
Output Output For each case output the area of their intersected part in a single line.accurate up to 2 decimal places.
Sample Input 1.00 1.00 3.00 3.00 2.00 2.00 4.00 4.00 5.00 5.00 13.00 13.00 4.00 4.00 12.50 12.50
Sample Output 1.00 56.25
import java.util.Scanner;public class Main{ public static void main(String[] args) { Scanner cin = new Scanner(System.in); while (cin.hasNext()) { double t; double sum = 0.0; double x1 = cin.nextDouble(); double y1 = cin.nextDouble(); double x2 = cin.nextDouble(); double y2 = cin.nextDouble(); double x3 = cin.nextDouble(); double y3 = cin.nextDouble(); double x4 = cin.nextDouble(); double y4 = cin.nextDouble(); if (x1 > x2) { t = x1; x1 = x2; x2 = t; } if (y1 > y2) { t = y1; y1 = y2; y2 = t; } if (x3 > x4) { t = x3; x3 = x4; x4 = t; } if (y3 > y4) { t = y3; y3 = y4; y4 = t; } x1 = max(x1, x3); x2 = min(x2, x4); y1 = max(y1, y3); y2 = min(y2, y4); if (x1 > x2 || y1 > y2) { System.out.println("0.00"); } else { sum = (x2 - x1) * (y2 - y1); System.out.printf("%.2f", sum); System.out.println(); } } } private static double min(double x2, double x4) { if (x2 < x4) return x2; else return x4; } private static double max(double x1, double x3) { if (x1 > x3) return x1; else return x3; }}新闻热点
疑难解答
